3.343 \(\int \frac{B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=62 \[ \frac{(B+2 C) \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{(B-C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

((B + 2*C)*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) + ((B - C)*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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Rubi [A]  time = 0.073853, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {4052, 12, 3794} \[ \frac{(B+2 C) \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{(B-C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^2,x]

[Out]

((B + 2*C)*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) + ((B - C)*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

Rule 4052

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] +
Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*(2*m + 1) + (b*B*(m + 1) - a*(A*(m + 1) - C*
m))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=\frac{(B-C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \frac{a (B+2 C) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac{(B-C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{(B+2 C) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a}\\ &=\frac{(B-C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{(B+2 C) \tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.287011, size = 46, normalized size = 0.74 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \left ((C-B) \sec ^2\left (\frac{1}{2} (c+d x)\right )+2 (2 B+C)\right )}{6 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^2,x]

[Out]

((2*(2*B + C) + (-B + C)*Sec[(c + d*x)/2]^2)*Tan[(c + d*x)/2])/(6*a^2*d)

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Maple [A]  time = 0.052, size = 60, normalized size = 1. \begin{align*}{\frac{1}{2\,d{a}^{2}} \left ( -{\frac{B}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+B\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +C\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)

[Out]

1/2/d/a^2*(-1/3*tan(1/2*d*x+1/2*c)^3*B+1/3*C*tan(1/2*d*x+1/2*c)^3+B*tan(1/2*d*x+1/2*c)+C*tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 0.947883, size = 126, normalized size = 2.03 \begin{align*} \frac{\frac{C{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}} + \frac{B{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(C*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 + B*(3*sin(d*x + c)/(cos(
d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d

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Fricas [A]  time = 0.466989, size = 144, normalized size = 2.32 \begin{align*} \frac{{\left ({\left (2 \, B + C\right )} \cos \left (d x + c\right ) + B + 2 \, C\right )} \sin \left (d x + c\right )}{3 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*((2*B + C)*cos(d*x + c) + B + 2*C)*sin(d*x + c)/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{B \sec{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(B*sec(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**2/(sec(c + d*x)
**2 + 2*sec(c + d*x) + 1), x))/a**2

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Giac [A]  time = 1.1259, size = 81, normalized size = 1.31 \begin{align*} -\frac{B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{6 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(B*tan(1/2*d*x + 1/2*c)^3 - C*tan(1/2*d*x + 1/2*c)^3 - 3*B*tan(1/2*d*x + 1/2*c) - 3*C*tan(1/2*d*x + 1/2*c
))/(a^2*d)